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3 years ago

Which of the following chemical equations is unbalanced?

1 answer:

7 0

Answer is A.
There are 4 hydrogen on LHS while there are 6 on RHS.
The correct reaction is
N2 + 3H2 2NH3
This is known as Haber’s process which is used to make Ammonia(NH3).
Hope this helps :)

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3 years ago

In a single wire, how much current would be required to generate 1 Tesla magnetic field at a 2 meter distance away from the wire

aev [14]

Answer:

12.56 A.

Explanation:

The magnetic field of a conductor carrying current is give as

H = I/2πr ............................... Equation 1

Where H = Magnetic Field, I = current, r = distance, and π = pie

Making I the subject of the equation,

I = 2πrH............... Equation 2

Given: H = 1 T, r = 2 m.

Constant: π = 3.14

Substitute into equation 2

I = 2×3.14×2×1

I = 12.56 A.

Hence, the magnetic field = 12.56 A.

5 0

3 years ago

Three identical resistors are connected in parallel. The equivalent resistance increases by 630 when one resistor is removed and

strojnjashka [21]

Answer:

each resistor is 540 Ω

Explanation:

Let's assign the letter R to the resistance of the three resistors involved in this problem. So, to start with, the three resistors are placed in parallel, which results in an equivalent resistance $R_e$ defined by the formula:

$\frac{1}{R_e}=\frac{1}{R} } +\frac{1}{R} } +\frac{1}{R} \\\frac{1}{R_e}=\frac{3}{R} \\R_e=\frac{R}{3}$

Therefore, R/3 is the equivalent resistance of the initial circuit.

In the second circuit, two of the resistors are in parallel, so they are equivalent to:

$\frac{1}{R'_e}=\frac{1}{R} +\frac{1}{R}\\\frac{1}{R'_e}=\frac{2}{R} \\R'_e=\frac{R}{2} \\$

and when this is combined with the third resistor in series, the equivalent resistance ( $R''_e$ ) of this new circuit becomes the addition of the above calculated resistance plus the resistor R (because these are connected in series):

$R''_e=R'_e+R\\R''_e=\frac{R}{2} +R\\R''_e=\frac{3R}{2}$

The problem states that the difference between the equivalent resistances in both circuits is given by:

$R''_e=R_e+630 \,\Omega$

so, we can replace our found values for the equivalent resistors (which are both in terms of R) and solve for R in this last equation:

$\frac{3R}{2} =\frac{R}{3} +630\,\Omega\\\frac{3R}{2} -\frac{R}{3} = 630\,\Omega\\\frac{7R}{6} = 630\,\Omega\\\\R=\frac{6}{7} *630\,\Omega\\R=540\,\Omega$

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